I am playing with the simplest of polynomial vector spaces - the Legendre polynomials (I hope I have that name right! :-) where $\langle P,Q\rangle = \int_{-1}^{+1}P(x)Q(x)dx$

This is straightforward as long as I restrict myself to real numbers.

But the first axiom of an inner product is that $\langle A,B\rangle = \operatorname{conjugate}(\langle B,A\rangle)$. I have verified (by pen & paper - quaint fellow, eh? :-) that this works nicely if my vectors are n-tuples of complex numbers over the field of $ℂ$, $ℂ^n/ℂ$. But when my vectors are polynomials [with real and complex coefficients] how do I even define the conjugate of my vector? I have already determined that conjugating the complex coefficients does not produce a polynomial that satisfies that first axiom.

I may also be getting side-tracked and defining the conjugate of such a vector is not even a requirement for producing an operation that fits that axiom.

Course correction, please?

  • $\begingroup$ I think what you mean in your first paragraph is $P\cdot Q=\int_{-1}^1P(x)Q(x)\,dx$ (which is achieved by puuting dollar signs at both ends of P\cdot Q=\int_{-1}^1P(x)Q(x)\,dx), though I don't know what Legendre polynomials have to do with it. In the complex case, $\int_{-1}^1P\overline Q\,dx$ should work, where $\overline Q$ is what you get by taking the complex conjugate of the coefficients. If you think this doesn't work, please provide an example with the details of how you think it fails. $\endgroup$ – Gerry Myerson Dec 29 '14 at 19:29
  • 1
    $\begingroup$ Conjugation distributes over multiplication (Proof: $x = a + bi$, $x^* = a - bi$, $y = c + di$, $y^* = c - di$, $xy = ac + bdi^2 + (ad + bc)i = (ac - bd) + (ad + bc)i$ $x^*y^* = ac + (-1)^2bdi^2 + (-ad -bc)i = (xy)^*$.) so $(P(z))^* = P(z^*)$ for any polynomial $P$. As @GerryMyerson points out, $\int_{-1}^1 PQ^*dx$ will work; note that $\int_{-1}^1 PP^*dx$ boils down to $\int_{-1}^1 (\text{something positive in x}) dx$. $\endgroup$ – J. C. Salomon Dec 29 '14 at 22:39


Your straightforward assertion that "In the complex case, $P\cdot Q=\int_{-1}^1P(x)Q_*(x)\,dx$ should work, where $Q_*$ is what you get by taking the complex conjugate of the coefficients" annoyed me into intensely re-examining my Perl code and I discovered two glaring errors:

  1. The polynomial conjugate function was calculating the conjugate but returning the original object.

  2. The test program was expecting the original value instead of the conjugate.

So my errors, in months-old code, cancelled each other out. The new inner-product function I am writing did not fall into that trap and called me on in.

Bottom line: Indeed, Gerry, it now works so well that, in 64-bit FP arithmetic, there is a 100% match between $\int_{-1}^1P(x)\overline Q(x)\,dx$ and $(\int_{-1}^1Q(x)P(x)\,dx)_*$

(BTW I'm using $_*$ to indicate that last conjugate because I don't quite have the hang of the overline over the entire parenthesized expression.)

Thank you Gerry, for that "annoyance". And JCS for starting me on the proper syntax for the mathematical expressions in TeX commands.

-- JS


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